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seraphim [82]
3 years ago
12

The solution of 2x -3=7

Mathematics
2 answers:
sineoko [7]3 years ago
7 0
ANSWER:
2x-3=7
2x=7+3
2x=10
x=10/2
x=5
HOPE IT HELPS!!!!
PLEASE MARK BRAINLIEST!!!!
WARRIOR [948]3 years ago
4 0

Answer:x=5

Step-by-step explanation:

2x-3=7

2x=3+7

X=10/2

X=5

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What is the area of the figure below?<br><br> Pls answer fast
jonny [76]

Answer:

15

Step-by-step explanation:

area = lxw

3x5=15

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3 years ago
Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.
Elden [556K]

Answer:

The answer is "cirtical\  points \ (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})"

Step-by-step explanation:

Given:

g(x) = (x-e)^3e^{-(x-e)}

Find critical points:

g(x) = (x-e)^3e^{(e-x)}

differentiate the value with respect of x:

\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r}  \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]

critical points g'(x)=0

\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3

So,

The critical points of (x,g(x))\equiv  (e,0),(e+3,\frac{27}{e^3})

7 0
3 years ago
Consider a collection of envelopes consisting of 1 red envelope, 3 blue envelopes, 2 green envelopes, and 3 yellow envelopes if
nasty-shy [4]

Answer:

the probability that at least one envelope is a yellow envelope is 16/21

Step-by-step explanation:

The probability that at least one envelope is a yellow envelope is P(Y);

P(Y) = 1 - P(Y)'

P(Y)' is the probability that no envelope is a yellow envelope.

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red envelope = 1

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Since there is no replacement;

P(Y)' = 6/9 × 5/8 × 4/7

P(Y)' = 5/21

From equation 1;

P(Y) = 1 - 5/21

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the probability that at least one envelope is a yellow envelope is 16/21.

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4 years ago
Coyuntura by five from 135 to 175. Write these Numbers and describe the pattern
Iteru [2.4K]
You are going up by 5
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3 years ago
I think it can be a triangle but want another opinion
pashok25 [27]

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I think It's a triangle too

Step-by-step explanation:

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